What would be the partial derivative with respect to y for the following function: f(x,y)=x^2sin(3xy)?
I think I should use the product rule and treat x as a constant, but beyond that I am lost.
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2006-12-07 15:41:02 · 5 answers · asked by cheezo12 1 in Science & Mathematics ➔ Mathematics
To solve for the partial derivative with respect to y, treat x like a constant and then solve.
Therefore, the partial derivative of x^2(sin[3xy]) with respect to y would be
x^2(cos(3xy))(3x)
2006-12-07 15:44:03 · answer #1 · answered by Puggy 7 · 0⤊ 0⤋
no need for the use of the product rule here. just consider everything except for y to be a constant. so you get the partial of f with respect to y to be 3*x^3*cos(3xy).
2006-12-07 23:43:58 · answer #2 · answered by Quinn M 1 · 0⤊ 0⤋
There's no product to do. As you stated x is a constant.
Go ahead and pull out the constant
x^2 (d/dy sin(3xy)) = x^2 * 3x * cos(3xy)= 3x^3 cos(3xy)
Recall that
d/dx sin(f(x))= cos(f(x)) f'(x).
You shouldn't have been thinking product rule but rather chain rule.
2006-12-08 00:02:09 · answer #3 · answered by chris.blanton 2 · 0⤊ 0⤋
you were correct by treating x as a constant. you use substitution and make u=3xy, then your fcn is x^2cos(u)*(u'). then your answer should be x^2cos(3xy)*(3x).
2006-12-08 00:18:08 · answer #4 · answered by Jessie 2 · 0⤊ 0⤋
d/dy f(x,y)=x^(2)sin(3xy)
d/dy f(x,y)= 3x^(2) cos (3xy)
2006-12-08 00:03:58 · answer #5 · answered by rfdsp2003 1 · 0⤊ 0⤋